Question

For the reaction, $2NO\left(g\right)+O_2\left(g\right)\to 2N{O}_2\left(g\right)$ the value of $\Delta H$ is $-113.1\text{kJ}$. If $6.0\text{moles}$ of $NO$ reacts with $3.0\text{moles}$ of $O_2$ at $1.0\text{atm}$ and $300\text{K}$ to form $N{O}_2$. Calculate the work done against a pressure of $1.0\text{atm}$.
(Take $R=\frac{1}{12}{\text{L atm mol}}^{-1}{\text{K}}^{-1}$)

• A. $7.56\text{kJ}$
• B. $-75\text{J}$
• C. $-7.56\text{kJ}$
• D. $75\text{J}$

Answer 1

The correct option is A $7.56\text{kJ}$

The reaction is
$\begin{array}{cccccc}2NO\left(g\right)& +& O_2\left(g\right)& \to & 2N{O}_2\left(g\right);& \Delta H=-113.1\text{kJ}\\ \text{2 mol}& & \text{1 mol}& & \text{2 mol}& \to \left(1\right)\\ (3\times 2)\text{mol}& & (3\times 1)\text{mol}& & (3\times 2)\text{mol}& \Delta H=3\times (-113.1)\text{kJ}\end{array}$
We know that
$PV=nRT$
$\therefore V_{\text{reactants}}=\frac{nRT}{P}$
Here, $n=(6+3)\text{moles = 9 moles}$
$R=0.0821{\text{L atm mol}}^{-1}{\text{K}}^{-1}$
$T=298K$, $P=1.0$ atm
$\therefore V_{\text{reactants}}=\frac{9\times \frac{1}{12}\times 300}{1}=225\text{L}$

Also, $V_{\text{products}}=\frac{nRT}{P}$
$\therefore V_{\text{products}}=\frac{6\times \frac{1}{12}\times 300}{1}=150\text{L}$

Now, work done $\left(W\right)=-P\Delta V$
$=-1(150-225)=75\text{L atm}$
$=75\times 101.3\text{J}$ $(\because 1\text{L atm}=101.3\text{J})$
$=7.56\times {10}^3\text{J}=7.56\text{kJ}$